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If so, what is it?

P=V^2/R, so increase voltage and keep R low. R is the resistance of the load, which is equal to the internal resistance. (With superconducting internal and external load R=0 but we have a current density limit of superconductors and inductance. Power could then be transferred to an antenna or motor. I guess then we'd still have impedance matched load.)

Increasing dielectric thickness d allows for higher voltage, but we also need to increase insulation at the fringe to avoid corona discharge, which increases volume/mass by d^3, while power increases with d^2, so power density actually decreases.

Could a capacitor made of two charged hollow spheres in the vacuum of space be scaled to ever increasing power/mass (on even volume) density? Large spheres/distance would limit "corona" discharge.

Do we get peak power density by fully exploiting the highest dielectric strength material (diamond?), regardless of capacitor size?

I'm asking, because

https://de.wikipedia.org/wiki/Energiespeicher#Speichern_elektrischer_Energie

said the maximum power of a normal capacitor is 10 kW (0.01 MW in column "max. Leistung in MW"="max. power in MW"), which seemed wrong.

https://en.wikipedia.org/wiki/Power-to-weight_ratio#Electrostatic,_electrolytic_and_electrochemical_capacitors

mentions a 3300 V, 100 kA capacitor, which works out to 330 MW.

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  • \$\begingroup\$ \$P=V^2/R\$ is the formula for for power dissipated by a resistor. A capacitor is not a resistor, so this formula does not apply to a capacitor. One useful formula is \$E=\frac{1}{2}CV^2\$ which gives the energy stored in a capacitor. \$\endgroup\$ – The Photon Apr 23 at 21:23
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    \$\begingroup\$ When we say "power density" or "energy density" we usually mean power or energy per unit volume. Increasing the plate size may increase the energy stored in your capacitor, but it will also increase the physical volume, so it has no effect on energy density. \$\endgroup\$ – The Photon Apr 23 at 21:25
  • \$\begingroup\$ @ThePhoton R is, ofc, the load. For maximum power through the load R should be equal to internal resistance, I think. E=CV^2/2 is energy, but I'm asking about power (density). \$\endgroup\$ – darsie Apr 23 at 21:28
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    \$\begingroup\$ There is no "power density" in a capacitor as it doesn't store power. There is energy density (by volume or weight). I think your question is suffering some translation problems. Please edit your user profile to state your location or language (German?) and we will try to fix the question. \$\endgroup\$ – Transistor Apr 23 at 21:38
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    \$\begingroup\$ Are you looking to collide some atoms with a capacitor discharge into a cryogenic coil? Do you have any contraints? They do make plastic caps with 1 milliohm ESR capacitor rated at 1kV. Did you want a bigger example? \$\endgroup\$ – Sunnyskyguy EE75 Apr 24 at 22:54
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Your post is unclear so I just offer the following observations.

Wikipedia's Power to weight ratio mentions a 3300 V, 100 kA capacitor, which works out to 330 MW.

That is in the section Electrostatic, electrolytic and electrochemical capacitors and the details are:

Type                      General Atomics 3330CMX2205 High Voltage Capacitor
Capacity                  20.5 mF
Voltage                   3300 V
Energy to weight ration   2.3 kJ/kg
Power to weight ratio     6.8 MW/kg @ 100 kA

330 MW is not "the power of the capacitor". It is the peak power the capacitor can deliver.

We can work out roughly how long this will last as follows:

$$ Q = CV $$ $$ V = \frac {Q}{C} $$ $$ \frac {dV}{dt} = \frac {dQ}{dt}\frac {1}{C} = \frac {I}{C} = \frac {100k}{20.5m} = 4.8 \ \text {MV/s} $$

Your capacitor, if it discharged linearly would be flat in \$ \frac {3300}{4.8M} = 0.7 \ \text {ms} \$.

However, I suspect that 100 kA can only be delivered into a short circuit and in that case the output voltage is zero so the power delivered to the load is zero and all the energy is dissipated in the internal resistance of the capacitor. This is probably the main flaw with your thinking. If at 3300 V the device can supply 100 kA max then it implies the internal resistance is \$ \frac {V}{I} = \frac {3300}{100k} = 33 \ \text m \Omega \$.

Now, using your maximum power transfer idea we would use a 33 mΩ load, current would be reduced to 50 kA, voltage reduced to half of 3300 and the maximum instantaneous power to the load would be \$ 50k \times 3.3k /2 = 82.5 \ \text {MW} \$.

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    \$\begingroup\$ Nope. I'd tell you why. \$\endgroup\$ – Transistor Apr 23 at 22:10
  • \$\begingroup\$ Do you agree with my answer? \$\endgroup\$ – Sunnyskyguy EE75 Apr 23 at 22:11
  • \$\begingroup\$ I don't really understand the question so I'm not sure what the correct answer is but I don't know why you're discussing self-discharge. \$\endgroup\$ – Transistor Apr 23 at 22:16
  • \$\begingroup\$ Linear discharge is unrealistic for RC. RC takes infinite time to discharge and thus deliver 0 average power. I do mean peak power. True, maximum power does half the voltage, so that must be our goal. The listed case says "6.8 MW/kg @ 100 kA". We could look up capacitor mass and get actual power etc. I guess it's power at rated current, but it's not important for our discussion. \$\endgroup\$ – darsie Apr 23 at 22:25
  • \$\begingroup\$ The reason you asked was in reference to a Wiki page on Energy storage and vacuum capacitors with a Dk=1 are not a viable option. Peak Discharge current is not a measure of energy storage. \$\endgroup\$ – Sunnyskyguy EE75 Apr 23 at 22:28
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If you are referring to peak pulsed power then the limit is pretty high even for small capacitors. The limiting factors are the rated voltage and impedance vs. frequency.

As a rough estimate a 1206 size 200V 10nF ceramic capacitor (C1206C103K2GAC) or similar has an impedance of of only 25mOhm at 45MHz. (See http://ksim.kemet.com/).

200V ^ 2/ 25mOhm = 1.6MW peak @ 45MHz

Of course most of that impedance is ESR so the capacitor would explode immediately (as capacitors are prone to do when shorted).

How much peak power the capacitor can safely do on a repetitive basis can be determined from the allowed ripple current vs temperature rise graph. In this case its 6.67A (so 1.3kW peak).

Larger capacitors could obviously have higher peaks.

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  • \$\begingroup\$ How does this demonstrate the correct Capacitor peak power density? It only disproves the bad assumption using ESR or Isc. What is your answer? Power-to-weight ratio for batteries is therefore less meaningful without reference to corresponding energy-to-weight ratio and cell temperature, using Peukert's law. -1 \$\endgroup\$ – Sunnyskyguy EE75 Apr 24 at 20:11
  • \$\begingroup\$ @SunnyskyguyEE75 I think the OP was only asking about peak wattage at some time instant without regard to the total amount of energy or energy density. Weather peak wattage is useful by itself is another matter. The only thing that limits peak power should be the complex impedance of the capacitor and the voltage across it at some time instant. As for determining a bound for all capacitors, I am not sure there is a practically determinable upper limit. My aim was only to show that values quoting hundreds of kilowatts or megawatts were not unreasonable. \$\endgroup\$ – user4574 Apr 24 at 22:05

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